closedness of subshift Consider the set $\sum_n^{+} = \{0,1,2,,,n-1\}^{\mathbb{N}}$ consisting of all one-sided sequence of elements in $\{0,1,2,...,n-1\}$. By Tychonoff's , this's compact (w.r.t product topology). Is there an alternate way to prove the compactness here?.
Define the shift map $\sigma : \{0,1,2,,,n-1\}^{\mathbb{N}} \to \{0,1,2,,,n-1\}^{\mathbb{N}}$ by $\sigma(x_o\,x_1\,x_2\,,...) = (x_1\,x_2\,,...)$, which is continuous.
Then, define a subshift of finite type $\sum_A \subset \sum_n^{+}$. How to prove this is closed as a subsapce of $(\sum_n^{+}, \sigma)$  ?
 A: 
Is there an alternate way to prove the compactness here?

Yes, there is a direct way. Put $\bar n=\{0,\dots, n-1\}$ and suppose to the contrary that there exists an open cover $\mathcal U$ of $X={\bar n}^{\Bbb N}=\sum_n^{+}$ which has no finite subcover. Then there exists a number $n_0\in \bar n$ such that $\mathcal U$ has no finite subcover of $X_0=\{(x_n)\in X: x_0=n_0\}$. By induction with respect to $k$ we can construct a sequence $(n_k)\in [n]^{\Bbb N}$  such that for each $k$,  $\mathcal U$ has no finite subcover of $X_k=\{(x_n)\in X: x_i=n_i$ for each $i=1,\dots, k\}$. There exists $U\in\mathcal U$ such that $(n_k)\in U$. Since the family $\{X_k:k\in\Bbb N\}$ is a base at the point $(x_n)$, there exists $k\in\Bbb N$ such $X_k\subset U$, a contradiction.
A: The Tikhonov theorem is the easiest way to prove that $\Sigma_n^+$ is compact, but you could also prove that it’s compact by embedding it in the Cantor set: the middle-thirds Cantor set is compact as a closed, bounded subset of $\Bbb R$, it’s not too hard to show that it’s homeomorphic to $\Sigma_2^+$ and that $\Sigma_2^+$ is homeomorphic to $\Sigma_{2^n}^+$ for each $n\in\Bbb N$, and it’s easy to embed $\Sigma_n^+$ in $\Sigma_m^+$ whenever $n\le m$.
I had to look up subshift of finite type, but if I understand correctly we can think of $A$ an an $n\times n$ binary matrix with entries $a_{i,j}$ for $i,j\in\{0,1,\ldots,n-1\}$, in which case
$$\Sigma_A^+=\{\langle x_k:k\in\Bbb N\rangle\in\Sigma_n^+:a_{x_k,x_{k+1}}=1\text{ for each }k\in\Bbb N\}\,.$$
In other words, $\Sigma_A^+$ consists of the sequences in $\Sigma_n^+$ that are walks along the graph with vertex set $\{0,1,\ldots,n-1\}$ and adjacency matrix $A$.
The most straightforward way to prove that $\Sigma_A^+$ is closed in $\Sigma_n^+$ is to show that its complement is open. Suppose that $x=\langle x_k:k\in\Bbb N\rangle\in\Sigma_n^+\setminus\Sigma_A^+$; then there is an $\ell\in\Bbb N$ such that $a_{x_\ell,x_{\ell+1}}=0$. Let
$$U=\{\langle y_k:k\in\Bbb N\rangle:y_\ell=x_\ell\text{ and }y_{\ell+1}=x_{\ell+1}\}\,;$$
then $U$ is a basic open set in the product space $\Sigma_n^+$, $x\in U$, and $U\cap\Sigma_A^+=\varnothing$, so $\Sigma_n^+\setminus\Sigma_A^+$ is open, and $\Sigma_A^+$ is closed.
